Question

You are booking tickets for a flight to Brazil, the Cessna airplane only has 10 tickets. It is common knowledge that only 85% of people who buy plane tickets actually make it to the gate, so the airline sells 11 tickets for your flight. What is the probability that there will be at least 1 empty seat?

Answer 1

The Cessna airplane has 10 seats.

Probability of a passenger showing up = 0.85

To have at least one empty seat, the aircraft should fly with less than 10 passenger

Probability that less than 10 passenger are coming = 1 – probability that 10 or more people coming.

The airline sells 11 tickets.

Hence the probability of airplane flying with at least 1 empty seat = 1 – (probability that 10 people showed up + probability that 11 people showed up)

We will denote probability of 10 people showed up as P(10) and probability of 11 people showed up as P(11)

Now

Probability that 10 people showed up:

Randomly selecting 10 out of 11 (sold tickets) multiplied with the probability of coming raised to power of number of people coming multiplied with probability of not coming raised to power of number of people not came.

Hence,

P(10) = ¹¹C₁₀ * 0.85^10 * 0.15^1 = 11 * 0.197 * 0.15 = 0.3248

Probability that 11 people showed up:

P(11) = ¹¹C₁₁ * 0.85^11 * 0.15^0 = 1 * 0.167 * 1 = 0.1673

Hence probability of 10 or more people showed = P(10) + P(11) = 0.3248 + 0.1673 = 0.4921

So, probability that the airplane will fly with at least one empty seat = 1 – (P(10) + P(11)) = 1 – 0.4921 = 0.5079

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